Chapter 7: Introduction to Trigonometry
Trigonometry — literally "triangle measurement" — is one of the oldest and most practical branches of mathematics. Ancient astronomers used it to map the stars, engineers use it to design bridges and buildings, and physicists use it to model waves and oscillations. In this chapter you will learn the foundations of trigonometry: how the ratios of sides in a right triangle give rise to the sine, cosine, and tangent functions; how the unit circle extends these functions to all angles; how radians provide an alternative and often more natural way to measure angles; and how to graph, transform, and invert trigonometric functions. By the end of this chapter you will be prepared for the more advanced identities, equations, and applications that appear in Precalculus and Calculus.
7.1 Right Triangle Trigonometry
Trigonometry begins with right triangles. Given a right triangle with an acute angle $\theta$, we label the three sides relative to that angle: the side directly across from $\theta$ is the opposite side, the side adjacent to $\theta$ (that is not the hypotenuse) is the adjacent side, and the longest side — opposite the right angle — is the hypotenuse.
Definition: The Three Primary Trig Ratios (SOH CAH TOA)
For an acute angle $\theta$ in a right triangle:
$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan\theta = \frac{\text{opposite}}{\text{adjacent}}$$
The mnemonic SOH CAH TOA encodes these: Sine = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, Tangent = Opposite / Adjacent.
These ratios depend only on the angle $\theta$, not on the size of the triangle. Any two right triangles that share the same acute angle are similar, so their side ratios are identical. This is the fundamental insight that makes trigonometry so powerful: knowing just one angle and one side of a right triangle is enough to determine all the other sides.
Finding Missing Sides
To find an unknown side of a right triangle, choose the trig ratio that relates the known side, the unknown side, and the given angle. Then solve the resulting equation.
Example 7.1.1 — Finding a Missing Side
Problem: In a right triangle, the angle $\theta = 35°$ and the hypotenuse is $20$. Find the length of the side opposite $\theta$.
Step 1 Identify the relationship. We know the hypotenuse and want the opposite side, so we use sine:
$$\sin 35° = \frac{\text{opposite}}{20}$$
Step 2 Solve for the opposite side:
$$\text{opposite} = 20 \sin 35° \approx 20 \times 0.5736 \approx 11.47$$
Finding Missing Angles
When you know two sides of a right triangle and need to find an acute angle, compute the appropriate trig ratio and then use the inverse function on your calculator. For instance, if $\sin\theta = 0.6$, then $\theta = \sin^{-1}(0.6) \approx 36.87°$.
Example 7.1.2 — Finding an Angle
Problem: A right triangle has legs of length $5$ and $12$. Find the measure of the angle opposite the side of length $5$.
Step 1 We know both legs. The side of length $5$ is opposite our target angle, and $12$ is adjacent. Use tangent:
$$\tan\theta = \frac{5}{12}$$
Step 2 Apply the inverse tangent:
$$\theta = \tan^{-1}\!\left(\frac{5}{12}\right) \approx 22.62°$$
Special Right Triangles
Two families of right triangles appear so frequently in mathematics that their side ratios should be memorized. They come from bisecting an equilateral triangle and from cutting a square along its diagonal.
The 45-45-90 Triangle
In a $45°$-$45°$-$90°$ triangle, the two legs are equal and the hypotenuse is $\sqrt{2}$ times as long as each leg. If each leg has length $a$, the sides are:
$$a : a : a\sqrt{2}$$
Therefore: $\sin 45° = \cos 45° = \dfrac{\sqrt{2}}{2}$ and $\tan 45° = 1$.
The 30-60-90 Triangle
In a $30°$-$60°$-$90°$ triangle, if the shorter leg (opposite $30°$) has length $a$, then the longer leg (opposite $60°$) has length $a\sqrt{3}$ and the hypotenuse has length $2a$:
$$a : a\sqrt{3} : 2a$$
Therefore: $\sin 30° = \dfrac{1}{2}$, $\cos 30° = \dfrac{\sqrt{3}}{2}$, $\tan 30° = \dfrac{\sqrt{3}}{3}$, $\sin 60° = \dfrac{\sqrt{3}}{2}$, $\cos 60° = \dfrac{1}{2}$, $\tan 60° = \sqrt{3}$.
Example 7.1.3 — Using Special Triangles
Problem: A ladder leans against a wall making a $60°$ angle with the ground. The foot of the ladder is $4$ meters from the base of the wall. How long is the ladder?
Step 1 The $60°$ angle is between the ground and the ladder. The ground distance ($4$ m) is the side adjacent to the $60°$ angle. The ladder is the hypotenuse.
Step 2 Use cosine:
$$\cos 60° = \frac{4}{\text{hypotenuse}} \implies \frac{1}{2} = \frac{4}{\text{hyp}} \implies \text{hyp} = 8 \text{ meters}$$
Study Tip: Memorize the side ratios of the 30-60-90 and 45-45-90 triangles. These ratios recur throughout trigonometry, standardized tests, and college-level math. Being able to recall them instantly will save you considerable time.
7.2 The Unit Circle
Right triangle trigonometry only covers acute angles ($0° < \theta < 90°$). To extend trigonometric functions to all angles — including obtuse angles, negative angles, and angles greater than $360°$ — we use the unit circle, which is the circle of radius $1$ centered at the origin: $x^2 + y^2 = 1$.
Definition: Trig Functions on the Unit Circle
Place an angle $\theta$ in standard position: vertex at the origin, initial side along the positive $x$-axis. The terminal side of $\theta$ intersects the unit circle at a point $P(x, y)$. We define:
$$\cos\theta = x \qquad \sin\theta = y \qquad \tan\theta = \frac{y}{x} \; (x \neq 0)$$
This agrees with SOH CAH TOA for acute angles (where $\text{hyp} = 1$ on the unit circle) and extends naturally to all angles.
From the equation $x^2 + y^2 = 1$ and the definitions above, we immediately obtain the most important identity in trigonometry:
Pythagorean Identity
$$\sin^2\theta + \cos^2\theta = 1$$
This holds for every angle $\theta$. It is the trigonometric version of the Pythagorean Theorem.
Reference Angles
A reference angle is the acute angle formed between the terminal side of an angle and the $x$-axis. It always lies between $0°$ and $90°$. To evaluate trig functions at any angle, find its reference angle, look up the trig value of that reference angle, and then apply the correct sign based on the quadrant.
- Quadrant I: reference angle $\alpha = \theta$
- Quadrant II: $\alpha = 180° - \theta$
- Quadrant III: $\alpha = \theta - 180°$
- Quadrant IV: $\alpha = 360° - \theta$
Quadrant Signs: All Students Take Calculus
The signs of sine, cosine, and tangent depend on which quadrant the terminal side falls in. Since $\cos\theta = x$ and $\sin\theta = y$, the signs follow from whether $x$ and $y$ are positive or negative in each quadrant.
| Quadrant | $\sin\theta$ ($y$) | $\cos\theta$ ($x$) | $\tan\theta$ ($y/x$) | Mnemonic |
|---|---|---|---|---|
| I | $+$ | $+$ | $+$ | All positive |
| II | $+$ | $-$ | $-$ | Sine positive |
| III | $-$ | $-$ | $+$ | Tangent positive |
| IV | $-$ | $+$ | $-$ | Cosine positive |
The mnemonic "All Students Take Calculus" reads counterclockwise from Quadrant I: All, Sine, Tangent, Cosine. Each word tells you which trig function(s) are positive in that quadrant.
Key Coordinates on the Unit Circle
Combining the special triangle values with the reference angle technique gives us exact coordinates for every multiple of $30°$ and $45°$ around the unit circle. Here are the values in the first quadrant; all other quadrants follow by adjusting signs.
| Angle | $0°$ | $30°$ | $45°$ | $60°$ | $90°$ |
|---|---|---|---|---|---|
| $\cos\theta$ | $1$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{1}{2}$ | $0$ |
| $\sin\theta$ | $0$ | $\dfrac{1}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $1$ |
Example 7.2.1 — Evaluating Trig Functions Using the Unit Circle
Problem: Find $\sin 225°$ and $\cos 225°$.
Step 1 $225°$ is in Quadrant III. Reference angle: $225° - 180° = 45°$.
Step 2 In Quadrant III, both sine and cosine are negative.
Step 3 Apply the reference angle values:
$$\sin 225° = -\sin 45° = -\frac{\sqrt{2}}{2} \qquad \cos 225° = -\cos 45° = -\frac{\sqrt{2}}{2}$$
Example 7.2.2 — Finding Trig Values from One Known Value
Problem: If $\sin\theta = \dfrac{3}{5}$ and $\theta$ is in Quadrant II, find $\cos\theta$ and $\tan\theta$.
Step 1 Use the Pythagorean identity:
$$\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$$
Step 2 In Quadrant II, cosine is negative: $\cos\theta = -\dfrac{4}{5}$.
Step 3 Compute tangent:
$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{-4/5} = -\frac{3}{4}$$
Interactive: The Unit Circle
Drag the slider below to move a point around the unit circle. Watch how the coordinates $(\cos\theta, \sin\theta)$ change as the angle $\theta$ varies.
The point traces the unit circle as the angle slider changes. The dashed lines show the sine and cosine projections.
7.3 Radian Measure
Degrees are intuitive but somewhat arbitrary — why divide a circle into $360$ parts? Radians offer a more natural and mathematically elegant way to measure angles. In calculus and higher mathematics, radians are the standard unit because they simplify many formulas.
Definition: Radian
One radian is the measure of a central angle that intercepts an arc whose length equals the radius of the circle. A full circle has circumference $2\pi r$, so a full rotation is $2\pi$ radians.
Converting Between Degrees and Radians
Since $360° = 2\pi$ radians, we have $180° = \pi$ radians. This gives the conversion factors:
Degree-Radian Conversion
$$\text{Degrees to radians: } \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180}$$
$$\text{Radians to degrees: } \theta_{\text{deg}} = \theta_{\text{rad}} \times \frac{180}{\pi}$$
Example 7.3.1 — Converting Between Units
Problem: (a) Convert $150°$ to radians. (b) Convert $\dfrac{7\pi}{6}$ radians to degrees.
(a) $150° \times \dfrac{\pi}{180} = \dfrac{150\pi}{180} = \dfrac{5\pi}{6}$ radians.
(b) $\dfrac{7\pi}{6} \times \dfrac{180}{\pi} = \dfrac{7 \times 180}{6} = \dfrac{1260}{6} = 210°$.
Here are the most commonly encountered degree-radian equivalences:
| Degrees | $0°$ | $30°$ | $45°$ | $60°$ | $90°$ | $120°$ | $180°$ | $270°$ | $360°$ |
|---|---|---|---|---|---|---|---|---|---|
| Radians | $0$ | $\dfrac{\pi}{6}$ | $\dfrac{\pi}{4}$ | $\dfrac{\pi}{3}$ | $\dfrac{\pi}{2}$ | $\dfrac{2\pi}{3}$ | $\pi$ | $\dfrac{3\pi}{2}$ | $2\pi$ |
Arc Length
The relationship between an arc, its central angle, and the radius is beautifully simple when the angle is measured in radians.
Arc Length Formula
If a central angle $\theta$ (in radians) subtends an arc of length $s$ on a circle of radius $r$, then:
$$s = r\theta$$
This formula is the reason radians are preferred in science and engineering. If we used degrees, the formula would need an extra factor of $\frac{\pi}{180}$, making it less clean.
Example 7.3.2 — Arc Length
Problem: A circle has radius $10$ cm. Find the length of the arc intercepted by a central angle of $\dfrac{3\pi}{4}$ radians.
$$s = r\theta = 10 \times \frac{3\pi}{4} = \frac{30\pi}{4} = \frac{15\pi}{2} \approx 23.56 \text{ cm}$$
Sector Area
A sector is the "pie slice" region bounded by two radii and an arc. Its area is a fraction of the full circle's area, proportional to the central angle.
Sector Area Formula
The area of a sector with central angle $\theta$ (in radians) and radius $r$ is:
$$A = \frac{1}{2}r^2\theta$$
Example 7.3.3 — Sector Area
Problem: A sprinkler waters a circular sector with radius $8$ meters and central angle $120°$. Find the area of the watered region.
Step 1 Convert to radians: $120° \times \frac{\pi}{180} = \frac{2\pi}{3}$.
Step 2 Apply the formula:
$$A = \frac{1}{2}(8)^2 \left(\frac{2\pi}{3}\right) = \frac{1}{2}(64)\left(\frac{2\pi}{3}\right) = \frac{64\pi}{3} \approx 67.02 \text{ m}^2$$
Study Tip: Always check that your angle is in radians before using $s = r\theta$ or $A = \frac{1}{2}r^2\theta$. If the angle is given in degrees, convert it first. Using degrees directly in these formulas is the single most common error in this section.
7.4 Graphing Sine and Cosine
The graphs of $y = \sin x$ and $y = \cos x$ are smooth, continuous waves that repeat every $2\pi$ units. These wave patterns model countless real-world phenomena: sound waves, alternating current, tides, pendulum motion, and seasonal temperature cycles. Learning to read and transform these graphs is one of the most important skills in Algebra 2.
The Basic Graphs
The graph of $y = \sin x$ starts at the origin, rises to a peak of $1$ at $x = \frac{\pi}{2}$, returns to $0$ at $x = \pi$, drops to a trough of $-1$ at $x = \frac{3\pi}{2}$, and returns to $0$ at $x = 2\pi$ to complete one full cycle. The graph of $y = \cos x$ has the same shape but starts at its peak: it begins at $1$, drops to $0$ at $\frac{\pi}{2}$, reaches $-1$ at $\pi$, returns to $0$ at $\frac{3\pi}{2}$, and returns to $1$ at $2\pi$.
Key properties shared by both basic graphs:
- Domain: $(-\infty, \infty)$
- Range: $[-1, 1]$
- Period: $2\pi$ (the horizontal length of one complete cycle)
- Amplitude: $1$ (the distance from the midline to a peak or trough)
- Midline: $y = 0$ (the horizontal line halfway between the max and min)
The General Sinusoidal Form
The general form of a transformed sinusoidal function is:
$$y = A\sin\bigl(B(x - C)\bigr) + D$$
(The same form applies with cosine.) Each of the four parameters $A$, $B$, $C$, $D$ controls a specific transformation:
| Parameter | Name | Effect | Formula |
|---|---|---|---|
| $A$ | Amplitude | Vertical stretch or compression. If $A < 0$, the graph reflects over the midline. | Amplitude $= |A|$ |
| $B$ | Frequency factor | Horizontal stretch or compression, changing the period. | Period $= \dfrac{2\pi}{|B|}$ |
| $C$ | Phase shift | Horizontal translation. Positive $C$ shifts right; negative shifts left. | Phase shift $= C$ |
| $D$ | Vertical shift | Moves the midline up ($D > 0$) or down ($D < 0$). | Midline: $y = D$ |
Common mistake: The argument must be factored as $B(x - C)$, not left as $Bx - C$. For instance, $y = \sin(2x - \pi)$ must be rewritten as $y = \sin\bigl(2(x - \frac{\pi}{2})\bigr)$ before reading off the phase shift. The phase shift is $\frac{\pi}{2}$ to the right, not $\pi$.
Example 7.4.1 — Identifying Parameters
Problem: For $y = -3\cos(4x + \pi) + 2$, identify the amplitude, period, phase shift, and midline.
Step 1 Factor the argument: $4x + \pi = 4\bigl(x + \frac{\pi}{4}\bigr) = 4\bigl(x - (-\frac{\pi}{4})\bigr)$.
So $A = -3$, $B = 4$, $C = -\frac{\pi}{4}$, $D = 2$.
Step 2 Compute:
- Amplitude: $|A| = 3$
- Period: $\frac{2\pi}{4} = \frac{\pi}{2}$
- Phase shift: $\frac{\pi}{4}$ to the left
- Midline: $y = 2$
- The negative $A$ means the cosine wave is reflected vertically (starts going down from the midline instead of up).
Example 7.4.2 — Writing an Equation from a Description
Problem: A sinusoidal function has amplitude $5$, period $6\pi$, phase shift $\frac{\pi}{3}$ to the right, and midline $y = -1$. Write a sine equation.
Step 1 $A = 5$, $C = \frac{\pi}{3}$, $D = -1$.
Step 2 Find $B$: Period $= \frac{2\pi}{|B|} = 6\pi \implies |B| = \frac{2\pi}{6\pi} = \frac{1}{3}$.
Step 3 Assemble:
$$y = 5\sin\!\left(\frac{1}{3}\left(x - \frac{\pi}{3}\right)\right) - 1$$
Interactive: Sine/Cosine Transformer
Use the sliders below to adjust $A$, $B$, $C$, and $D$ and watch how the graph of $y = A\sin(B(x - C)) + D$ (blue) transforms compared to the base graph $y = \sin x$ (dashed gray). The dashed green line shows the midline $y = D$.
Drag the sliders to explore amplitude, period, phase shift, and vertical shift in real time.
7.5 Graphing Tangent and Other Trig Functions
The tangent, cotangent, secant, and cosecant functions have graphs that look very different from the smooth waves of sine and cosine. Their defining feature is the presence of vertical asymptotes where the function is undefined.
The Tangent Function
Since $\tan x = \frac{\sin x}{\cos x}$, the function is undefined wherever $\cos x = 0$, which occurs at $x = \frac{\pi}{2} + n\pi$ for any integer $n$. At these values the graph has vertical asymptotes.
Properties of $y = \tan x$
Domain: All real numbers except $x = \frac{\pi}{2} + n\pi$
Range: $(-\infty, \infty)$
Period: $\pi$ (half the period of sine and cosine)
No amplitude (the function is unbounded)
The graph passes through the origin, is increasing on each branch, and has an inflection point at each $x$-intercept.
For a transformed tangent $y = A\tan(Bx)$, the period becomes $\frac{\pi}{|B|}$ and the vertical stretch factor is $|A|$. Vertical asymptotes shift according to the transformation.
Example 7.5.1 — Graphing a Transformed Tangent
Problem: Find the period and two consecutive asymptotes of $y = \tan(3x)$, and sketch one period.
Step 1 Period $= \frac{\pi}{|B|} = \frac{\pi}{3}$.
Step 2 Asymptotes occur where $3x = \frac{\pi}{2} + n\pi$, i.e., $x = \frac{\pi}{6} + \frac{n\pi}{3}$.
Two consecutive asymptotes: $x = -\frac{\pi}{6}$ (when $n = -1$) and $x = \frac{\pi}{6}$ (when $n = 0$).
Step 3 Key points within this period:
- $x = -\frac{\pi}{12}$: $y = \tan\!\left(-\frac{\pi}{4}\right) = -1$
- $x = 0$: $y = \tan(0) = 0$
- $x = \frac{\pi}{12}$: $y = \tan\!\left(\frac{\pi}{4}\right) = 1$
The graph rises from $-\infty$ to $+\infty$ between the two asymptotes, passing through these three points.
Cotangent, Secant, and Cosecant
The remaining three trig functions are reciprocals of the primary three:
- $\cot x = \frac{\cos x}{\sin x} = \frac{1}{\tan x}$: undefined at $x = n\pi$, period $\pi$, decreasing on each branch.
- $\sec x = \frac{1}{\cos x}$: undefined at $x = \frac{\pi}{2} + n\pi$, period $2\pi$, range $(-\infty, -1] \cup [1, \infty)$.
- $\csc x = \frac{1}{\sin x}$: undefined at $x = n\pi$, period $2\pi$, range $(-\infty, -1] \cup [1, \infty)$.
| Function | Period | Asymptotes | Range |
|---|---|---|---|
| $\tan x$ | $\pi$ | $x = \frac{\pi}{2} + n\pi$ | $(-\infty, \infty)$ |
| $\cot x$ | $\pi$ | $x = n\pi$ | $(-\infty, \infty)$ |
| $\sec x$ | $2\pi$ | $x = \frac{\pi}{2} + n\pi$ | $(-\infty, -1] \cup [1, \infty)$ |
| $\csc x$ | $2\pi$ | $x = n\pi$ | $(-\infty, -1] \cup [1, \infty)$ |
Graphing Tip: To graph $\csc x$ or $\sec x$, first sketch the corresponding reciprocal function ($\sin x$ or $\cos x$) as a dashed guide curve. Wherever the guide curve crosses the $x$-axis, draw a vertical asymptote. The branches of the reciprocal function open away from the $x$-axis, touching the guide curve at its peaks and troughs.
Example 7.5.2 — Period and Asymptotes of Cosecant
Problem: Find the period and vertical asymptotes of $y = \csc(2x)$.
Step 1 The period of $\csc(Bx)$ is $\frac{2\pi}{|B|} = \frac{2\pi}{2} = \pi$.
Step 2 Vertical asymptotes occur where $\sin(2x) = 0$, i.e., $2x = n\pi$, so $x = \frac{n\pi}{2}$.
The asymptotes are at $x = 0, \pm\frac{\pi}{2}, \pm\pi, \pm\frac{3\pi}{2}, \ldots$
7.6 Inverse Trigonometric Functions
Trigonometric functions are periodic, so they fail the Horizontal Line Test on their full domains and are not one-to-one. To define inverse functions, we must restrict each trig function to an interval where it is one-to-one. The resulting inverses "undo" the trig functions and are essential for solving trigonometric equations.
The Three Principal Inverse Trig Functions
Arcsine: $y = \arcsin x = \sin^{-1} x$ means $\sin y = x$ with $-\dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$.
Domain: $[-1, 1]$. Range: $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.
Arccosine: $y = \arccos x = \cos^{-1} x$ means $\cos y = x$ with $0 \leq y \leq \pi$.
Domain: $[-1, 1]$. Range: $[0, \pi]$.
Arctangent: $y = \arctan x = \tan^{-1} x$ means $\tan y = x$ with $-\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$.
Domain: $(-\infty, \infty)$. Range: $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$.
Notation caution: The notation $\sin^{-1}x$ means the inverse sine function (arcsine), not $\frac{1}{\sin x}$. To write the reciprocal, use $(\sin x)^{-1}$ or $\csc x$. This notational trap catches many students.
Evaluating Inverse Trig Functions
To evaluate an expression like $\arcsin\!\left(\frac{1}{2}\right)$, ask: "What angle $\theta$ in the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ satisfies $\sin\theta = \frac{1}{2}$?" Since $\sin\frac{\pi}{6} = \frac{1}{2}$ and $\frac{\pi}{6}$ is in the required range, the answer is $\frac{\pi}{6}$ (or equivalently, $30°$).
Example 7.6.1 — Evaluating Inverse Trig Expressions
Problem: Find the exact values of: (a) $\arccos\!\left(-\dfrac{\sqrt{2}}{2}\right)$, (b) $\arctan(1)$, (c) $\arcsin(-1)$.
(a) We need $\theta \in [0, \pi]$ with $\cos\theta = -\frac{\sqrt{2}}{2}$. The reference angle is $\frac{\pi}{4}$ (since $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$). Negative cosine means Quadrant II: $\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
(b) We need $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with $\tan\theta = 1$. Since $\tan\frac{\pi}{4} = 1$, the answer is $\frac{\pi}{4}$.
(c) We need $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with $\sin\theta = -1$. Since $\sin\!\left(-\frac{\pi}{2}\right) = -1$, the answer is $-\frac{\pi}{2}$.
Compositions of Trig and Inverse Trig Functions
A common type of problem asks you to evaluate a trig function of an inverse trig function, such as $\cos(\arcsin x)$. The strategy is to let $\theta$ equal the inner inverse trig expression, draw a right triangle, label the sides, and then read off the desired trig value.
Example 7.6.2 — Evaluating a Composition
Problem: Find $\cos\!\left(\arctan\dfrac{5}{12}\right)$.
Step 1 Let $\theta = \arctan\frac{5}{12}$. Then $\tan\theta = \frac{5}{12}$ with $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Since the value is positive, $\theta$ is in Quadrant I.
Step 2 In the right triangle for $\theta$: opposite $= 5$, adjacent $= 12$. By the Pythagorean Theorem, hypotenuse $= \sqrt{25 + 144} = \sqrt{169} = 13$.
Step 3 Read off cosine:
$$\cos\!\left(\arctan\frac{5}{12}\right) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{12}{13}$$
Example 7.6.3 — Another Composition
Problem: Find $\sin\!\left(\arccos\!\left(-\dfrac{3}{5}\right)\right)$.
Step 1 Let $\theta = \arccos\!\left(-\frac{3}{5}\right)$. Then $\cos\theta = -\frac{3}{5}$ with $\theta \in [0, \pi]$. Since cosine is negative, $\theta$ is in Quadrant II, where sine is positive.
Step 2 Use the Pythagorean identity:
$$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$
We take the positive root because $\theta$ is in Quadrant II.
Study Tip: When evaluating compositions, the "triangle method" always works. Let $\theta$ equal the inner inverse trig expression, draw a right triangle with the appropriate sides, find the missing side using the Pythagorean Theorem, and then read off whatever trig ratio the outer function asks for. Pay attention to the quadrant to get the sign right.
7.7 Practice Problems
Test your understanding of Chapter 7 with these problems. Click "Show Solution" to reveal a full worked solution for each one.
Problem 1
In a right triangle, one acute angle is $40°$ and the adjacent side is $15$. Find the opposite side.
Show Solution
We know the adjacent side and want the opposite side, so we use tangent:
$$\tan 40° = \frac{\text{opposite}}{15} \implies \text{opposite} = 15\tan 40° \approx 15 \times 0.8391 \approx 12.59$$
Problem 2
A 45-45-90 triangle has a hypotenuse of length $10\sqrt{2}$. Find the length of each leg.
Show Solution
In a 45-45-90 triangle, the sides are in the ratio $a : a : a\sqrt{2}$. So the hypotenuse $= a\sqrt{2}$.
$$a\sqrt{2} = 10\sqrt{2} \implies a = 10$$
Each leg has length $10$.
Problem 3
Find the exact values of $\sin 315°$ and $\cos 315°$ using the unit circle.
Show Solution
$315°$ is in Quadrant IV. Reference angle: $360° - 315° = 45°$.
In Quadrant IV, cosine is positive and sine is negative:
$$\sin 315° = -\sin 45° = -\frac{\sqrt{2}}{2}$$
$$\cos 315° = +\cos 45° = \frac{\sqrt{2}}{2}$$
Problem 4
If $\cos\theta = -\dfrac{5}{13}$ and $\theta$ is in Quadrant III, find $\sin\theta$ and $\tan\theta$.
Show Solution
Use $\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$.
In Quadrant III, sine is negative: $\sin\theta = -\frac{12}{13}$.
$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-12/13}{-5/13} = \frac{12}{5}$$
(Tangent is positive in Quadrant III, as expected.)
Problem 5
Convert $\dfrac{5\pi}{12}$ radians to degrees, and convert $225°$ to radians.
Show Solution
Radians to degrees: $\dfrac{5\pi}{12} \times \dfrac{180}{\pi} = \dfrac{5 \times 180}{12} = \dfrac{900}{12} = 75°$.
Degrees to radians: $225° \times \dfrac{\pi}{180} = \dfrac{225\pi}{180} = \dfrac{5\pi}{4}$.
Problem 6
A circle has radius $6$ cm. Find the arc length and sector area for a central angle of $\dfrac{2\pi}{3}$ radians.
Show Solution
Arc length: $s = r\theta = 6 \times \frac{2\pi}{3} = 4\pi \approx 12.57$ cm.
Sector area: $A = \frac{1}{2}r^2\theta = \frac{1}{2}(36)\left(\frac{2\pi}{3}\right) = 12\pi \approx 37.70$ cm$^2$.
Problem 7
Find the amplitude, period, phase shift, and midline of $y = 4\sin\!\left(2x - \dfrac{\pi}{3}\right) - 1$.
Show Solution
Rewrite: $y = 4\sin\!\left(2\!\left(x - \frac{\pi}{6}\right)\right) - 1$.
Parameters: $A = 4$, $B = 2$, $C = \frac{\pi}{6}$, $D = -1$.
- Amplitude: $|A| = 4$
- Period: $\frac{2\pi}{2} = \pi$
- Phase shift: $\frac{\pi}{6}$ to the right
- Midline: $y = -1$
Problem 8
Write a cosine equation with amplitude $2$, period $8\pi$, phase shift $\pi$ to the right, and midline $y = 3$.
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$A = 2$, $C = \pi$, $D = 3$.
Period $= \frac{2\pi}{|B|} = 8\pi \implies |B| = \frac{2\pi}{8\pi} = \frac{1}{4}$.
$$y = 2\cos\!\left(\frac{1}{4}(x - \pi)\right) + 3$$
Problem 9
Find the period and two consecutive vertical asymptotes of $y = \tan\!\left(\dfrac{\pi x}{2}\right)$.
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Period $= \frac{\pi}{|B|} = \frac{\pi}{\pi/2} = 2$.
Asymptotes occur where $\frac{\pi x}{2} = \frac{\pi}{2} + n\pi$, i.e., $x = 1 + 2n$.
Two consecutive asymptotes: $x = -1$ (when $n = -1$) and $x = 1$ (when $n = 0$).
Problem 10
Evaluate: (a) $\arcsin\!\left(\dfrac{\sqrt{3}}{2}\right)$, (b) $\arccos(0)$, (c) $\arctan(-\sqrt{3})$.
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(a) We need $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with $\sin\theta = \frac{\sqrt{3}}{2}$. Answer: $\theta = \frac{\pi}{3}$ (i.e., $60°$).
(b) We need $\theta \in [0, \pi]$ with $\cos\theta = 0$. Answer: $\theta = \frac{\pi}{2}$ (i.e., $90°$).
(c) We need $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with $\tan\theta = -\sqrt{3}$. The reference angle is $\frac{\pi}{3}$ (since $\tan\frac{\pi}{3} = \sqrt{3}$). The negative sign places us in the fourth-quadrant portion of the range: $\theta = -\frac{\pi}{3}$.
Problem 11
Find the exact value of $\sin\!\left(\arccos\dfrac{4}{5}\right)$.
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Let $\theta = \arccos\frac{4}{5}$. Then $\cos\theta = \frac{4}{5}$ with $\theta \in [0, \pi]$. Since $\cos\theta > 0$, $\theta$ is in Quadrant I where sine is positive.
$$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$$
Problem 12
A Ferris wheel has a diameter of $50$ meters with its center $30$ meters above the ground. It completes one revolution every $4$ minutes. Write a sinusoidal equation for the height $h(t)$ of a rider (in meters) as a function of time $t$ (in minutes), assuming the rider starts at the lowest point.
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The radius (amplitude) is $A = 25$. The midline (center height) is $D = 30$. The period is $4$ minutes, so $B = \frac{2\pi}{4} = \frac{\pi}{2}$.
Starting at the bottom means starting at the minimum. A negative cosine starts at its minimum:
$$h(t) = -25\cos\!\left(\frac{\pi}{2}\,t\right) + 30$$
Check: At $t = 0$: $h(0) = -25(1) + 30 = 5$ m (bottom, since center is 30 m and radius is 25 m). At $t = 2$: $h(2) = -25(-1) + 30 = 55$ m (top). Correct.